Thể tích O2 cần để đốt cháy hết 6,8 gam NH3 tạo thành khí N2 là

Hướng dẫn giải

Đáp án đúng là: B

$\begin{array}{l}{\mathrm{n}}_{{\mathrm{NH}}_3}=\frac{6,8}{17}=0,4\text{ mol}\\ 4{\mathrm{NH}}_3+3{\mathrm{O}}_2\stackrel{{\mathrm{t}}^{\mathrm{o}}}{\to } 2{\mathrm{N}}_2+6{\mathrm{H}}_2\mathrm{O}\\ \text{0,4}\to \text{ 0,3\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}mol}\\ \Rightarrow {\text{V}}_{{\mathrm{O}}_2}=24,79.0,3=\boxed{7,437\text{ (L)}}\end{array}$